### 3.139 $$\int \text{sech}(a+b x) \text{sech}(c+b x) \, dx$$

Optimal. Leaf size=36 $\frac{\text{csch}(a-c) \log (\cosh (a+b x))}{b}-\frac{\text{csch}(a-c) \log (\cosh (b x+c))}{b}$

[Out]

(Csch[a - c]*Log[Cosh[a + b*x]])/b - (Csch[a - c]*Log[Cosh[c + b*x]])/b

________________________________________________________________________________________

Rubi [A]  time = 0.0246153, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {5644, 3475} $\frac{\text{csch}(a-c) \log (\cosh (a+b x))}{b}-\frac{\text{csch}(a-c) \log (\cosh (b x+c))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]*Sech[c + b*x],x]

[Out]

(Csch[a - c]*Log[Cosh[a + b*x]])/b - (Csch[a - c]*Log[Cosh[c + b*x]])/b

Rule 5644

Int[Sech[(a_.) + (b_.)*(x_)]*Sech[(c_) + (d_.)*(x_)], x_Symbol] :> -Dist[Csch[(b*c - a*d)/d], Int[Tanh[a + b*x
], x], x] + Dist[Csch[(b*c - a*d)/b], Int[Tanh[c + d*x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2 - d^2, 0]
&& NeQ[b*c - a*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \text{sech}(a+b x) \text{sech}(c+b x) \, dx &=\text{csch}(a-c) \int \tanh (a+b x) \, dx-\text{csch}(a-c) \int \tanh (c+b x) \, dx\\ &=\frac{\text{csch}(a-c) \log (\cosh (a+b x))}{b}-\frac{\text{csch}(a-c) \log (\cosh (c+b x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.206235, size = 27, normalized size = 0.75 $\frac{\text{csch}(a-c) (\log (\cosh (a+b x))-\log (\cosh (b x+c)))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[a + b*x]*Sech[c + b*x],x]

[Out]

(Csch[a - c]*(Log[Cosh[a + b*x]] - Log[Cosh[c + b*x]]))/b

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 77, normalized size = 2.1 \begin{align*} -2\,{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,a-2\,c}} \right ){{\rm e}^{a+c}}}{b \left ({{\rm e}^{2\,a}}-{{\rm e}^{2\,c}} \right ) }}+2\,{\frac{\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ){{\rm e}^{a+c}}}{b \left ({{\rm e}^{2\,a}}-{{\rm e}^{2\,c}} \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)*sech(b*x+c),x)

[Out]

-2/b/(exp(2*a)-exp(2*c))*ln(exp(2*b*x+2*a)+exp(2*a-2*c))*exp(a+c)+2/b/(exp(2*a)-exp(2*c))*ln(1+exp(2*b*x+2*a))
*exp(a+c)

________________________________________________________________________________________

Maxima [A]  time = 1.6957, size = 92, normalized size = 2.56 \begin{align*} \frac{2 \, e^{\left (a + c\right )} \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b{\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )}} - \frac{2 \, e^{\left (a + c\right )} \log \left (e^{\left (-2 \, b x\right )} + e^{\left (2 \, c\right )}\right )}{b{\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)*sech(b*x+c),x, algorithm="maxima")

[Out]

2*e^(a + c)*log(e^(-2*b*x - 2*a) + 1)/(b*(e^(2*a) - e^(2*c))) - 2*e^(a + c)*log(e^(-2*b*x) + e^(2*c))/(b*(e^(2
*a) - e^(2*c)))

________________________________________________________________________________________

Fricas [B]  time = 1.87307, size = 490, normalized size = 13.61 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (-a + c\right ) - \sinh \left (-a + c\right )\right )} \log \left (\frac{2 \,{\left (\cosh \left (b x + c\right ) \cosh \left (-a + c\right ) - \sinh \left (b x + c\right ) \sinh \left (-a + c\right )\right )}}{\cosh \left (b x + c\right ) \cosh \left (-a + c\right ) -{\left (\cosh \left (-a + c\right ) + \sinh \left (-a + c\right )\right )} \sinh \left (b x + c\right ) + \cosh \left (b x + c\right ) \sinh \left (-a + c\right )}\right ) -{\left (\cosh \left (-a + c\right ) - \sinh \left (-a + c\right )\right )} \log \left (\frac{2 \, \cosh \left (b x + c\right )}{\cosh \left (b x + c\right ) - \sinh \left (b x + c\right )}\right )\right )}}{b \cosh \left (-a + c\right )^{2} - 2 \, b \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) + b \sinh \left (-a + c\right )^{2} - b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)*sech(b*x+c),x, algorithm="fricas")

[Out]

2*((cosh(-a + c) - sinh(-a + c))*log(2*(cosh(b*x + c)*cosh(-a + c) - sinh(b*x + c)*sinh(-a + c))/(cosh(b*x + c
)*cosh(-a + c) - (cosh(-a + c) + sinh(-a + c))*sinh(b*x + c) + cosh(b*x + c)*sinh(-a + c))) - (cosh(-a + c) -
sinh(-a + c))*log(2*cosh(b*x + c)/(cosh(b*x + c) - sinh(b*x + c))))/(b*cosh(-a + c)^2 - 2*b*cosh(-a + c)*sinh(
-a + c) + b*sinh(-a + c)^2 - b)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}{\left (a + b x \right )} \operatorname{sech}{\left (b x + c \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)*sech(b*x+c),x)

[Out]

Integral(sech(a + b*x)*sech(b*x + c), x)

________________________________________________________________________________________

Giac [B]  time = 1.22999, size = 135, normalized size = 3.75 \begin{align*} \frac{2 \, e^{\left (a + c\right )} \log \left (\frac{{\left | -{\left | e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )} \right |} + 2 \, e^{\left (2 \, b x + 2 \, a + 2 \, c\right )} + e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )} \right |}}{{\left | e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )} \right |} + 2 \, e^{\left (2 \, b x + 2 \, a + 2 \, c\right )} + e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}}\right )}{b{\left | e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )} \right |}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)*sech(b*x+c),x, algorithm="giac")

[Out]

2*e^(a + c)*log(abs(-abs(e^(2*a) - e^(2*c)) + 2*e^(2*b*x + 2*a + 2*c) + e^(2*a) + e^(2*c))/(abs(e^(2*a) - e^(2
*c)) + 2*e^(2*b*x + 2*a + 2*c) + e^(2*a) + e^(2*c)))/(b*abs(e^(2*a) - e^(2*c)))